$f(x) = 2x + 1$ for all real numbers. What is $f^{-1}(x)$, the inverse of $f(x)$ ? $2$ $4$ $6$ $8$ $\llap{-}4$ $\llap{-}6$ $\llap{-}8$ $2$ $4$ $6$ $8$ $\llap{-}4$ $\llap{-}6$ $\llap{-}8$ $f(x)$ $f^{-1}(x)$
Answer: $y = f(x)$ , so solving for $x$ in terms of $y$ gives $x=f^{-1}(y)$ $f(x) = y = 2x+1$ $y-1 = 2x$ $\dfrac{y}{2}-\frac{1}{2} = x$ $x = \dfrac{y}{2}-\frac{1}{2}$ So we know: $f^{-1}(y) = \dfrac{y}{2}-\frac{1}{2}$ Rename $y$ to $x$ $f^{-1}(x) = \dfrac{x}{2}-\frac{1}{2}$ Notice that $f^{-1}(x)$ is just $f(x)$ reflected across the line $y=x$.